Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is there a function (non piece-wise unlike below) which is discontinuous but has directional derivative at particular point? I have a manual that says the function has directional derivative at $(0,0)$ but is not continuous at $(0,0)$. $$f(x,y) = \begin{cases} \frac{xy^2}{x^2+y^4} & \text{ if } x \neq 0\\ 0 & \text{ if } x= 0 \end{cases}$$

Can anyone give me few examples which is not defined piece wise as above?

share|improve this question
    
Before I think about $|\cdot |$, does it count as a piece wise function? –  Git Gud Jan 16 '13 at 18:58
    
what would that be? could you give me link about |.| –  Santosh Linkha Jan 16 '13 at 18:59
    
I mean the absolute value. Of what? Yet to be determined, maybe $|x|$ or $|xy|$.. –  Git Gud Jan 16 '13 at 19:00
    
I think not .. i just don't want that condition imposed like f(x) = this when x = this or x=that –  Santosh Linkha Jan 16 '13 at 19:01
    
It's probably continuous anyway. –  Git Gud Jan 16 '13 at 19:02

3 Answers 3

up vote 1 down vote accepted

$$f(x,y)=\lim_{u\to0}\frac{xy^2+u^2}{x^2+y^4+u^2}$$

share|improve this answer
    
clever manipulation!! –  Santosh Linkha Jan 16 '13 at 19:36
    
@experimentX And I know you can't disallow limits if you would be willing to accept $e^x$ or $\sin x$. :) –  Hagen von Eitzen Jan 16 '13 at 19:37

The standard "elementary" functions are always continuous where they are defined, so this would be hard to do. You might try $$ f(x,y) = \arg( -\exp(i(y-x^2)(2x^2-y))) - (y-x^2)(2x^2-y) $$ where arg is the "principal branch" of the argument.

share|improve this answer

$f(x,y)=\frac{xy}{x^2+y^2}$ at $(x,y)\neq (0,0)$ and $=0$ at $(x,y)=(0,0)$

share|improve this answer
1  
OP asked for the function to not bet piece wise defined. –  Git Gud Jan 16 '13 at 18:55
    
also I am looking specifically (if possible) not piecewise functions with conditions –  Santosh Linkha Jan 16 '13 at 18:56
    
Since $f(x,x)\to\frac12$ when $x\to0$, some functional derivatives of $f$ at $(0,0)$ do not exist. –  Did Jan 16 '13 at 19:08

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.